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3 years ago

How many grams of zinc phosphate are formed when 48.1 mL of 2.18 M zinc nitrate reacts with excess potassium phosphate?

Chemistry

1 answer:

Roman55 [17]3 years ago

6 0

Answer:

15.4 g of Zn₃(PO₄)₂ are produced

Explanation:

Given data:

Mass of zinc phosphate formed = ?

Volume of zinc nitrate = 48.1 mL (0.05 L)

Molarity of zinc nitrate = 2.18 M

Solution:

Chemical equation:

3Zn(NO₃)₂ + 2K₃PO₄ → Zn₃(PO₄)₂ + 6KNO₃

Moles of zinc nitrate:

Molarity = number of moles / volume in litter

Number of moles = 2.18 M × 0.05 L

Number of moles = 0.109 mol

Now we will compare the moles of zinc phosphate with zinc nitrate from balanced chemical equation:

Zn(NO₃)₂ : Zn₃(PO₄)₂

3 : 1

0.109 : 1/3×0.109 = 0.04 mol

0.04 moles of Zn₃(PO₄)₂ are produced.

Mass of Zn₃(PO₄)₂:

Mass = number of moles × molar mass

Mass = 0.04 mol × 386.1 g/mol

Mass = 15.4 g

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If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

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But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

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